Graph y=(2x^2)/(x^2-1)

Find where the expression ${\displaystyle \frac{2x^2}{(x+1)(x-1)}}$ is undefined.

Since ${\displaystyle \frac{2x^2}{(x+1)(x-1)}}$${\displaystyle \to }$${\displaystyle \infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle -1}$ from the left and ${\displaystyle \frac{2x^2}{(x+1)(x-1)}}$${\displaystyle \to }$${\displaystyle -\infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle -1}$ from the right, then ${\displaystyle x=-1}$ is a vertical asymptote.

Since ${\displaystyle \frac{2x^2}{(x+1)(x-1)}}$${\displaystyle \to }$${\displaystyle -\infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle 1}$ from the left and ${\displaystyle \frac{2x^2}{(x+1)(x-1)}}$${\displaystyle \to }$${\displaystyle \infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle 1}$ from the right, then ${\displaystyle x=1}$ is a vertical asymptote.

List all of the vertical asymptotes:

Consider the rational function ${\displaystyle R\left(x\right)=\frac{a{x}^n}{b{x}^m}}$ where ${\displaystyle n}$ is the degree of the numerator and ${\displaystyle m}$ is the degree of the denominator.

Find ${\displaystyle n}$ and ${\displaystyle m}$.

Since ${\displaystyle n=m}$, the horizontal asymptote is the line ${\displaystyle y=\frac{a}{b}}$ where ${\displaystyle a=2}$ and ${\displaystyle b=1}$.

There is no oblique asymptote because the degree of the numerator is less than or equal to the degree of the denominator.

This is the set of all asymptotes.